Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 17 - Exercises and Problems - Page 326: 69

Answer

The proof is below.

Work Step by Step

We know the following equations: $\beta = \frac{1}{V}\frac{dV}{dT}$ $\alpha = \frac{1}{S}\frac{dS}{dT}$ We take the derivative of the equation for volume to obtain: $\frac{dV}{dT}=3S^2\frac{dS}{dT}$ $\frac{dV}{dT}=\frac{3V}{S}\frac{dS}{dT}$ $\frac{1}{V}\frac{dV}{dT}=\frac{3}{S}\frac{dS}{dT}$ Substituting the upper two equations, we find: $\beta = 3 \alpha$
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